浪潮杯第九届山东acm程序设计赛C题 Cities

C.Cities

题目描述

There are n cities in Byteland, and the i city has a value a . The cost of building a bidirectional road between two cities is the sum of their values. Please calculate the minimum cost of connecting these cities, which means any two cities can reach each other.

输入描述：

The first line is an integer T（T<= 1e5）
representing the number of test cases.

For each test case, the first line is an integer n(n <= 1e5), representing the number of cities, the
second line are n positive integers a (a <= 1e5),
representing their values.

输出描述：

For each test case, output an integer, which is the
minimum cost of connecting these cities.

示例1

输入

2

4

1 2 3 4

1

1

输出

12

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0

/*题意: Byteland有n个城市，i城市有价值a。 在两个城市之间建立双向道路的成本是它们的价值的总和。 请计算连接这些城市的最低成本，这意味着任何两个城市都可以相互联系*/

思路:

第一眼以为最小生成树，看了点数10的5次方，需要求出每个边的值，边树大约有10的9次方个，然后prim算法进行寻找，数据量太大，定会超时。。。。。

由此想到贪心算法：

想要将任意两个城市连接且道路花费之和最小，那可使每条道路的花费最少，道路的花费等于两端城市value之和，由此可知，只要每个城市与最小value的那个城市相连通，所得的花费必定是最小的。

因此，将最小value的城市放于中间与其他城市一一相连。

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#define p 100001
using namespace std;
int main(){
    int t;
    int a[p];
    scanf("%d",&t);
    while(t--){
        int n;
        scanf("%d",&n);
        for(int i=0;i<n;i++){
            scanf("%d",&a[i]);
        }
         if(n==1){
            printf("0\n");
            continue;
        }
        sort(a,a+n);
        int minn=a[0];
        long long sum=0;
        for(int i=1;i<n;i++){
            sum=sum+a[i]+minn;
        }
        printf("%lld\n",sum);
    }
}